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Print all the words after a match word

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# 1  
Old 05-15-2010
Print all the words after a match word
Hi,

I want to print all words till the last word after the match of "ERROR" word. For e.g.

I'll get an sqlplus error with
e.g. 1
Code:
 
$ ./calltest_fn.ksh
var test_var:=test_fn1; calltest_fn.ksh file1 file2 file3 ERROR at line 4: ORA-06550: line 4, column 11: PLS-00201: identifier 'test_fn1' must be declared ORA-06550: line 4, column 1: PL/SQL: Statement ignored

Now I want to print all the words after "ERROR" word. i.e. the output should be:
Code:
 
ERROR at line 4: ORA-06550: line 4, column 11: PLS-00201: identifier 'test_fn1' must be declared ORA-06550: line 4, column 1: PL/SQL: Statement ignored

Also e.g. 1 is one line only not separate lines; I know that we can print lines matching regEx with sed but don't how to print words?

-dips
# 2  
Old 05-15-2010
try this..
Code:
sed 's/\(.*\)\(ERROR.*\)/\2/g' filename

# 3  
Old 05-15-2010
Code:
sed "s/.*ERROR/ERROR/"

# 4  
Old 05-15-2010
Code:
sed 's/^var.*\(ERROR.*\)$/\1/' error

# 5  
Old 05-15-2010
If the word "ERROR" can itself appear in the error message, then the greedy nature of all the preceding solutions will only return part of the message. It may very well be that this is unlikely and of no concern. But, if it is, the following alternatives behave properly under such circumstances (the sh solution assumes that the error message is stored in e):
Code:
[ -z "${e##*ERROR*}" ] && echo "ERROR${e#*ERROR}"

Code:
sed 's/ERROR.*/&&/; s/.*\(ERROR.*\)\1/\1/'

Code:
awk 'i=index($0, "ERROR") {print substr($0, i)}'

Code:
perl -lne 'print $1 if /.*?(ERROR.*)$/'

Regards,
Alister
Last edited by alister; 05-15-2010 at 10:00 PM..
# 6  
Old 05-15-2010
All good and well, but it's an Oracle error, and that being superiorly efficient never feels the need to print the word ERROR more than once Smilie

But just to be sure:
Code:
sed "s/.*ERROR at line/ERROR at line/"

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